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Chapter five Volumetric analysis Principle Volumetric analysis is one of the most useful and accurate analytical techniques especially used for smaller a mounts (i.e in millmoles) of substance to be analyzed. It is based on the measurement of the volume of the added solution of known concentration which is sufficient to react with all of the analyzed substance in the sample then the concentration of the substance in the sample can be calculated Titrant: known solution which is gradually added which is often placed in a container knows as a buret Sample solution: solution which is to be analyzed is usually placed in conical flask Titration process by which the titrant is gradually added to sample solution Titration is stopped when the volume of titrant added is sufficient to react completely with the analyzed substance of the sample solution Equivalent point The point at which the volume of titrant added is sufficient to react completely with sample solution. Equivalent point: usually detected by an change in color of an indicator Indicator: chemical substance that its color change at equivalent point and it is added to the sample solution before starting the titration and this point is called the end point of titration.
However, the end point and the equivalent point are identical under ideal conditions. Types of titration and application principles with typical calculations of volumetric analysis operations are reviewed in this chapter. Starting point of volumetric analysis is preparation of Standard solution Solution which is used as titrant and whose concentration is accurately known. It is prepared by dissolving an accurately weighed quantity of highly pure material called as primary standard and diluting to an accurately known volume in a volumetric flask Standardization The standard solution is used to determine the exact concentration of any prepared solution Any titrant must be standardized before using in volumetric analysis measurements Back titrations Sometimes a sharp end point cannot be obtained and a slight excess of titrant is normally added to the sample solution in such cases back titration is used Back titration :- is used to determine the excess amount of the titrant by titration with another standard solution. From the amount of such a standard solution required for back titration the excess added amount of titrant can be computed.
Dilution of solutions It is frequently required to prepare a dilute solution of certain normality or molarity from a more concentrated one by qualitative dilution. Example ( 1 ) Calculate the volume of 0.25M of K2Cr2O7 solution required to prepare 500 ml of a 0.1 M solution MFinal × VFinal =Minitial ×Vinitial
or
Mf Vf = MiVi Vi=
=
= 200 ml
Example (2) Calculate the volume of 0.4M of Ba(OH)2 solution must be added to 50ml of 0.3M NaOH in order to prepare a solution of pH = 13.699 pOH= 14- 13.699 = 0.301 [OH-] = 0.5M x= ml of Ba(OH)2 mmol of OH- = 2(0.4)x + 50(0.3) = 15 + 0.8x MnO2 + 2Fe2+ + 4H+ → Mn2+ + 2Fe3+ + 2H2O The excess ferrous ion is determined by back titration with 0.02M of KMnO4 requiring 15ml according to the following reaction:5Fe2+ + MnO4- + 8H+ → Mn2+ + 5Fe3+ + 4H2O Calculate the manganese content in the sample as Mn3O4? excess Fe2+ used = 5 (MnO4- required in the 2 nd reaction) = 5(0.02×15)= 1.5 mmol total Fe2+ used = 50 ×0.1 = 5.0 mmol
net Fe2+reaction in 1st reaction= 5-1.5= 3.5 mmoles. MnO2 in the sample = × 3.5 = 1.75 mmoles MnO2 → Mn3O4 mmoles of Mn3O4 = (1.75) = 0.583 mmole wt% Mn3O4 =
× 100 =
×100 = 16.7%
Classification of Volumetric analysis 1. 2. 3. 4.
Neutralization method or (Acid – Base titrations) Precipitations titrations Oxidation – Reduction titrations Complexation titration
Neutralization method or (Acid – Base titrations) This type of volumetric analysis methods is based on the neutralization of hydrogen or hydroxide ion of acid and bases. The end point of such titrations are either detected by means of an indicators or by rapid change in pH which can be measured by the use of pH meter. Titration curves Graphs of pH versus the added volume of titrant. The inflection points in the titration curves are usually to be the end points of acid-base titrations. Atypical titration curve of strong acid with strong base is shown in figure the values have been computed for titration of 100 ml of 0.1 M HCl by 0.1 M NaOH It is shown that pH changes slowly at first until the equivalent point, arapid change in pH, there is nearly a vertical rise in the
region between pH 4-10, the pH changes slowly again after the equivalence point due to the addition of excess amount of NaOH, However, the magnitude of the vertical region depends on concentration of both the acid and base as shown in figure. Titration curve of strong acid-strong base Example (3) Calculate the pH and plot the resulting titration of 50ml of 0.1M HCl after the addition of 0, 10, 20, 40, 45, 48, 49, 50, 51, 55, 60, 80, and 100ml of 0.1M NaOH chemical reaction which takesplace is H+ + OH- ↔ H2O Initial conc. of [H+] in solution = 0.1M pH = 1 Initial mmoles of [H+] = 50 × 0.1= 5 mmoles After the addition of 10ml NaOH mmoles of OH-added = 10×0.1= 1.0 mmoles remaining mmoles of H+ = 5-1 =4 mmoles total volume of solution = 50 + 10 = 60 ml conc. of remaining H+ =
= 0.067M
pH= 1.76
and 50 on before the eq.point. ml, NaOH total volume mmoles remaining acid
[H+]
pH
0
50
5
0.1
1.0
10
60
4
0.0667
1.176
20
70
3
0.0429
1.368
40
90
1.0
0.0111
1.954
45
95
0.5
0.0053
2.279
48
98
0.2
0.0020
2.690
49
99
0.1
0.0010
2.996
Equivalence point: correspond to addition of 50ml of NaOH, [OH-] and pH can be calculated from the value of Kw Kw = [H+][OH-] = 1.0×10-14 at equivalence point [H+]=[OH-] = √
= 1×10-7M pH= 7
after the equivalence point, excess NaOH is added to the solution, pH of solution can be calculated from [H+],[OH-], pH, pOH, Kw pKw = pH + pOH = 14.0 the following table show the values of following mlNaOH total vol.ml mmol.excessOH- [OH-]
pOH
pH
51
101
0.1
9.9×10-4
3.004
10.996
55
105
0.5
4.77×10-3
2.322
11.672
60
110
1.0
9.09×10-3
2.041
11.95
80
130
3.0
0.0231
1.637
12.36
100
150
5.0
0.0333
1.477
12.58
The results of these calculations can be plotted as shown in this figure (2) Acid – Base Indicators Are highly colored organic dyes which exhibit a change in color when the pH of solution changed between certain limits, usually two pH units. This behavior can be explained from the fact that
most acid- base indicators are weak organic acids of general form HIn which can be dissociated as following HIn → H+ + Incolor1
color2
Ka = Hence, according to Henderson equation pH = pKa + log in practice the human eye sees color 1 when 2 when
=
and color
= 10, thus
when color 1 is seen pH = pKa + log
= pKa – 1
when color 2 is seen pH = pKa + log 10 = pKa + 1 pH = (pH)2 – (pH)1 = pKa+1 – pKa –(-1)= 2 which means a pH change of two units when the color is changed from color 1 to color 2. Midway in the titration [HIn]= [In-], hence the pKa of the indicator is close to pH at the equivalence point. However, a mixture of two or more indicators may be used in some cases to give better color change at the end point. Titration of weak acid with strong base or weak base with strong acid Titration of acetic acid with sodium hydroxid HOAC + NaOH → H2O + NaOAC Stage
Equation
Assumption
Before titration
[H+] =√
With titration
pH = pKa + log
At equivalent point
[OH-] = √
Ka <
[OAC-] > 100
[OH-] = [ excess titrant]
After equivalence point excess titrant Example (8)
Calculate the pH and plot the resulting titration curve for titration of 50ml of 0.1M HOAC by addition of 0, 10,25,50 and 60 ml of 0.1M NaOH Ka = 1.75× 10-5 x = [H+] = [OAC-]
At 0ml NaOH = 1.75× 10-5 [H+] = √
[ HOAC] = 0.1-x
Ka =√
0.1-x
pH = 2.88 after the addition of 10 ml NaOH mixture of weak acid + salt intial mmole of HOAC = 50 ×0.1 = 5 mmoles mmoles of [OH-] added = 10 ×0.1= 1 mmoles remaining mmoles of HOAC = 5-1 = 4 mmoles total volume = 50 +10 = 60 ml [ HOAC] =
= 0.0667 M
-3
= 1.32 ×10 M
0.1
[OAC-] =
= 0.0167 M
pH = pKa + log pH = - log 1.75 × 10-5 + log pH = 4.76 + 0.6 = 4.16 Similarly with addition of 25ml. [HOAC] =
[OAC-] =
= 0.0333 M
pH = 4.76 + log
= 0.0333M
= 4.76
At equil. Point = after the addition of 50 ml NaOH mmoles NaOH = 50 0.1 = 5 mmoles mmoles OAC- produced = 50 0.1 = 5 mmoles total volume = 100 ml [OAC-] = [OH-] = √
= 0.05M =√
= 5.35
10-6 M
pOH = 5.27 pH = 8.73 with the addition of 60 ml NaOH mmole = 60× 0.1 = 6 mmole NaOH total volume = 50 + 60 = 110 ml [OH-] =
= 0.00909M
pOH = 2.04
pH = 11.96
ml NaOH
pH
0
2.88
10
4.16
25
4.76
50
8.73t
60
11.96
Titration curve of weak base (NH3) with strong curve with reverse shape of titration curve reaction:NH3+ HCl → NH4Cl Stage
Equation
Assumption
Before titration
[OH-] =√
Kb <
With titration
pH = pKa + log
At equivalent point
[H+] = √
After equivalence
[NH4] > 100
[H+] = [ excess titrant]
Example (9) Calculate the pH and plot the resulting curve for the titration of 20 ml of 0.11M ammonia by the addition of 0, 5, 11, 15, 20, 22, 25, 30, 35, and 40ml of 0.1M HCl Kb = 1.79× 10-5 ? 1-
At 0 ml HCl
NH3 + H2O → NH4OH
x = equil. Conc [OH-] and [NH4+] 5
Kb = 1.79× 10- =
=
[OH-] =√
=√
pOH = -log [OH] = 2.85
= 1.4×10-3M pH = 11.15
Similar procedure of the previous example which give the following result and titration curve ml HCl
pH
0
11.15
5
9.79
11
9.29
15
8.9
20
8.3
22
5.27 eq. point
25
2.19
30
1.82
35
1.64
40
1.54
Precipitation titrations In this type of titration, titrant forms an insoluble product (precipitate) from its reaction with the sample solution such as: the titration of chloride ion with silver nitrate solution: end point of this type of titration can be detected either a) with an indicator or b) instrumental end point detection technique such as spectrophotometry. Two general methods are based on precipitation titrations 1. Mohr method
Silver nitrate solution is the most popular titrant for precipitation titrations:Ag+ + Cl- → AgCl (s) Ag+ + Br- → AgBr (s) Ag+ + I- → AgI (s) End point location:By the addition of sodium chromate, dark-orange silver chromate precipitate is formed upon the addition of excess Ag+ to the solution within a pH between 6-5 and 10-3 Reaction 2Ag+ + CrO42- → Ag2CrO4(s) 2. Volhard method This method can be used to analyze bromide and iodide solutions by the addition of an excess silver ion to precipitate silver halide. Excess Ag+ remaining in the solution after precipitation is back titrated with thiocyanate to yield silver thiocyante Ag+ + SCN- → AgSCN (s) End point location:Acidic solution of ferric ammonium sulfate as an indicator. Excess thiocyanate ion reacts with ferric ion to form a red complex of ferric thiocyanate :Fe3+ + SCN- → FeSCN2+ (red) Example A 50ml sample of bromide solution was analyzed by using volhard method. It is found that 10ml of 0.1M AgNO 3 is required, and back titrated with 0.0832M potassium thiocyanate. The end point of the back titration was reached after adding 5.34 ml of KCN solution. Calculate the bromide concentration (as molarity) in the sample solution? 1) Ag+ + Br- → AgBr (s)
2) Ag+ + SCN- → AgSCN (s)
mmoles of SCN- reacted in back titration reaction =5.34×0.0832= 0.444 mmoles = excess mmoles of Ag+ reacted total Ag+ reacted = 10(0.1) = 1.0 mmoles Ag+ reacted with bromide = 1-0.444 =0.556mmol of Br[Br-] =
=
= 0.01 M
Oxidation – Reduction titrations (Red.Ox) Oxidation- reduction reaction. A substance is oxidized when it loses electrons and is reduced when it gains electrons. In general redox reactions take place in two half reactions. In one half, oxidation takes place, and in other half, the reduction takes place Oxidation reaction
Fe2+ - e → Fe3+
reduction reaction
Ce4+ + e → Ce3+
over redox reaction
Fe2+ + Ce4+
Fe3+ + Ce3+
Chemical indicators: which are used to locate the end point of redox reactions which are organic compounds which have different colors in the oxidized form (In ) and the reduced form (Inr) of the indicator In + mH+ + ne- = Inr color 1
color 2
or
In + ne- = Inr color 1
color 2
m= can be either positive or negative several substance can be analyzed by redox titration. In most redox titration the end point can be located by direct titration with the titrant, However, in the some cases the back titration procedure is often used since an excess amount of titrant is required to force the slow reaction rapidly to completion
Iodometric titration Iodine ion is a weak reducing agent and when an excess of iodide is added to a solution of an oxidizing agent, I2 is produced in an amount equivalent to the oxidizing agent present Cr2O72- + 6I- + 14H+ ↔ 2Cr3+ + 3I2 + 7H2O The iodine is titrated with reducing agent, usually sodium thiosulfite (Na2S2O3) I2 + 2 S2O32- → 2I- + S4O62Analysis of oxidizing agent in this way is called "iodometric method" End point for iodometric titration is detected with starch- disappearance of the blue starch I2 color indicates the end of titration Example A metal sample containing copper with mass of 0.2514 was analyzed by idometric method. The sample was dissolved in acid and an excess a mounts of potassium iodide was added to the sample solution. The iodine produced was titrated with 26.32 ml of 0.10M sodium thiosulfate. Calculate the percent of copper in the metal sample? 2Cu2+ + 5I- → 2CuI + I3I3-+ 2S2O32- → 3I- + S4O62mmoles of thiosulfate = 26.3×0.1 = 2.63mmoles I3- produced in 1st reaction = 2.63 ( = 1.316 mmoles Copper ion reacted in 1st reaction = 1.316 × 2= 2.632mmoles mass of copper in the sample=2.632 mmol×
=167.2 mg
wt% Cu=
× 100=
4- Complexation titration Useful for determination of a large number of metals which from slightly dissociated complexes in solution containing complexing agent usually known as ligand which can be either a neutral molecule such as water or ammonia or an ion such as chloride cyanide or hydroxide. The complex can have either positive or negative charge, or it can be neutral. A formation constant (Kf) can be written for each step and for the overall reaction as shown in the following example NH3 + Ag+
Ag(NH3)+
Kf1 =
Ag(NH3)+
Ag(NH3)2+
Kf2 =
= 2×103 = 8×103
Over all reaction :Ag+ +2NH3
Ag(NH3)2+
Kf= Kf1. Kf2=
= 1.6×107
Complexation titration with EDTA EDTA (ethylene diamine tetra acetic acid) or its salt is the most titrants used as complexing agents about 95% of complexation titrations are carried out with EDTA which has the following chemical structures EDTA has four acidic dissociation constants since it is considered as tetraprotic acid with the symbol H4Y. It is dissociated by the following dissociation step:H4Y ↔ H+ + H3Y-
Ka1=
= 1×10-2
H3Y- ↔ H+ + H2Y2-
Ka2=
= 2.1×10-3
H2Y2- ↔ H+ + HY3-
Ka3=
= 6.92×10-7
HY3- ↔ H++ Y4-
= 5.5×10-11
Ka4=
Equilibrium calculations can be used to determine the fraction ( ) of each of the five forms of EDTA which is presented in solution at any pH for example the following equation can be used to calculate y4- at any pH =
= 1+
+
+
+
Where CT = total concentration of all forms of EDTA CT = [H4Y] + [H3Y-] + [H2Y2-] + [HY3-] + [Y4-] Similar equation can be used to calculate α HY3-, α H2Y2-, α H3Y- and H4YPlot of the fraction of each form of EDTA as function of pH are shown in the following figure (3) Example Calculate the fraction of EDTA that exists as Y4- at pH 10 ? at pH = 10 → [H+] = 10-10
=1+
+
+
+ = 1+
+
+
+
= 1+1.82 + 2.63 ×10-4+1.22×10-11+1.22×10-19 = 2.82 αY -4 = 0.355 Example
Use the result of the previous example to calculate the equilibrium concentration of ca2+ at pH = 10 if 100ml of solution of 0.1M Ca2+ is added to 100ml of 0.1M EDTA mmoles of Ca2+ added= 100 (0.1) = 10 mmoles mmoles of EDTA added= 100 (0.1) = 10 mmoles total volume = 100 +100 = 200 ml. initial conc. of CaY2- =
= 0.05M
the reaction of Ca2+ with EDTA can be represent by the equation Ca2+ + Y4- → CaY2x
x
0.05-x
0.05-x conc. of Ca2+ = [Y4-] = CT Y4-
x = equilibrium = 0.35 CT = 0.35x 3. Kf = x=√
=
=1×10-11
=1.2×10-6 M
Detection of the End point in complexation titration Spectrophotometry and several electroanytical techniques can be used to locate the end points of complexation titration. However, chemical indicators can be used as in other types of titrations. metallochromic indicators are widely used in complexation titration which from complexes with metal ions and exhibit different colors according to the following equation Inm ↔ HIn-1-m ↔ H2In2-m ↔ ………etc
Consequently, the pH of solution should be controlled when using metallochromic indicators.
Problems:1- The mohr method was used to determine sodium chloride content in a 1.0004 gm sample. The sample was dissolved in water and titrated to the end point with 32.36ml. of 0.1012M AgNO3. Calculate the wt% of NaCl in the sample? Ans : 19.13 wt% 2- A 25ml sample of potassium iodide was analyzed by using Volhard method. It is found that 20.0ml of 0.5015M AgNO3 is required, and the excess silver ion was back titrated with 32.35ml of 0.1038M KSCN solution. Calculate the concentration of KI solution? Ans :0.267M 3- A 0.5247gm metal sample containing copper was analyzed by iodometric method. The sample was dissolved in acid and diluted with distilled water to about 50ml. An excess amount of KI was added, and the liberated iodine was titrated with 34.87ml. of 0.1234M sodium thiosulfate. Calculate the percent of copper in the metal sample? Ans: 52.11% 4- Calculate the conditional formation constant (Kf) for the formation of Mn2+-EDTA complex at pH= 8 K1=1×10-2, K2=2.16×10-3, K3=6.92×10-7, K4=5.5×10-11? Ans: αY -4 = 5.4×10-3 , Kf = 2.05×1011 5- Calculate the conditional formation constant (Kf) for the formation of Fe2+-EDTA complex at pH= 9 K1=1×10-2, K2=2.16×10-3, K3=6.92×10-7, K4=5.5×10-11? Ans: αY -4 = 5.21×10-2 , Kf = 1.093×1013
6- Calculate [OH-] and the pH of the solution prepared by dissolving 7.82gm NaOH and 9.26gm Ba(OH)2 in water and diluting to 500ml? Ans : 0.67,13. 7- Calculate the volume of 0.1M H2SO4 must be added to 50ml of 0.10M NaOH solution to give a solution of 0.05M H2SO4 assuming additive volumes? Ans : 100ml 8- A 1.68 gm sample of iron ore is analyzed for iron content by dissolving in acid, coverting the iron to Fe2+ then titrating with 0.015M K2Cr2O7 solution according to the reaction:6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O If 35.6ml of potassium dichromate solution is required for titration, calculate the iron content in the sample expressed as wt% Fe2O3 Ans: 15. 9- A potassium permangate solution is prepared by dissolving 4.68gm KMnO4in water and diluting to 500ml. How many mill liters of this solution is required to react with iron in a 0.50 gm sample of iron ore containing 35.6 wt% Fe2O3? The reaction is 5Fe2+ + MnO4 + 8H+ → 5Fe3+ + Mn2+ + 4H2O Ans: 7.5ml 10- Calculate the pH for the titration of 50ml of 0.1M NaOH by the addition of 0, 10, 25, and 30ml of 0.2M HCl? Ans:13, 12, 70, 7.0, 1.90 11- Calculate the pH for the titration of 50ml of 0.1M NH3 by the addition of 0, 10, 25, 50 and 60ml of 0.1M HCl? Given that Kb =1.75×10-5 Ans: 11, 12, 9.85, 9.24, 5.27, 2.02